# StudySquare

# Methods of integration for OCR A-level Maths

This page covers the following topics:

1. Integration by parts

2. Integration by substitution

3. Integrating partial fractions

Integration by parts is a special method of integration which is used when two functions are being multiplied. They are defined as u and v accordingly, and the given formula is used to evaluate the interval.

Integration by substitution is a method to evaluate integrals by changing the variables. This method can be used when the integral is of the given form.

When the integrand is in the form of a proper fraction, the integral should be rewritten as partial fractions and then integrated using the given result.

# 1

Evaluate ∫1/(x + 5)dx.

∫1/(x + 5)dx = ln|x + 5| + c, where c is the constant of integration.

# 2

Use integration by parts to find ∫(5 + x)cos(x)dx.

Let u = 5 + x and dv = cos(x)dx.

Then, du = dx and v = sin(x).

∫ (5 + x)cos(x)dx = (5 + x)sin(x) − ∫sin(x)dx =

= (5 + x)sin(x) + cos(x) + c, where c is the constant of integration

(5 + x)sin(x) + cos(x) + c

# 3

Determine the following integral: ∫3x²sin(x³ − 10x) − 10sin(x³ − 10x)dx.

The integral can be rewritten as ∫(3x² − 10)sin(x³ − 10x)dx. To integrate by substitution, let u = x³ − 10x and du = (3x² − 10)dx. So, ∫(3x² − 10)sin(x³ − 10x)dx = ∫sin(u)du = −cos(u) + c = −cos(x³ − 10x) + c, where c is the constant of integration.

# 4

Evaluate ∫xeˣ.

Let u = x and dv = eˣdx. Then, du = 1dx and v = eˣ. So, ∫xeˣ = xeˣ − ∫eˣdx = xeˣ − eˣ + c, where c is the constant of integration.

# 5

Integrate the area of the given rectangle.

The integral is ∫15x²sin(8 − 5x³)dx. To integrate by substitution, let u = 8 − 5x³ and du = −15x²du. So, ∫15x²sin(8 − 5x³)dx = ∫−sin(u)dx = cos(u) + c = cos(8 − 5x³) + c, where c is the constant of integration.

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